Talk:Wilee5534
Q1. Your calculated energies for dimers and hydrogenated dimers are correct. The dimerization is under kinetic control, which you didn’t actually say, but I can surmise you came to this conclusion because of your discussion of transition states. Although you’re frontier molecular orbital interaction diagram shows the correct interaction (the secondary orbital interaction that stabilises the endo-TS wrt the exo-TS), your diagram could have been better labelled – it is clearer for Diels-Alder transition states to show both the orbital interactions that lead to bond formation and also the secondary interactions and label them as separate features. For the monohydrogenated endo-dimers bending strain is the major contributor to differences in energy as you say. These differences relate to the deviation from ideal sp2 bond angles of the remaining double bond after hydrogenation. It is correct that compound 4 is the thermodynamic product, but it is not possible to say which product is the kinetic product. You know that the endo-dimer produced in the first part of this question must be the kinetic product because it is the lower energy isomer and yet it is the major product. For the monohydrogenation, 4 is the product formed but it could be either the thermodynamic product OR the kinetic product. Alkene hydrogenation is usually carried out with a metal catalyst (e.g. palladium on charcoal) and hydrogen gas; under these conditions, reactions are usually under kinetic control (i.e. the products are not in equilibrium with the starting materials) and selective hydrogenation of alkenes is due to the steric effects that allow coordination of one alkene more easily than another.
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Q2. Your energy values are good and the reasoning for stereoselectivity in the nucleophilic additions is correct (assistance vs. repulsion). You could have included the jmols to show the structures of your lowest energy conformers and the CO dihedral angle, but overall this was well answered.
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Q3. It is correct that the cyclohexane unit has a large bearing on the energies for the atropisomers and you have found the conformation is different in both cases, but in both the images you have uploaded, the 6-ring is in the chair conformation. You should take care to upload the correct images and energy values (the numbers you got for species 10 are a little high – perhaps they relate to the chair structure presented). The alkene is specifically a “hyperstable alkene” and you are right in saying its stability relates to the instability of the alkane you would form by hydrogenating it.
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Q4. Your MOs look good, and as you say the distribution of the HOMO suggests that the alkene syn to the C-Cl bond is the better nucleophile. Likewise the values for the stretches are good and the jmols display this aspect well. You commented on an interaction between the pi bond of the anti double bond and the sigma* of the C-Cl bond when analysing the MOs. What implications does this kind of interaction have (especially for the C-Cl stretch and bond length)? This question could probably have done with a bit more discussion and a comparison to the monohydrogenated system. In that compound, with the anti-alkene removed, the pi-sigma* interaction is no longer possible which is why the C-Cl bond is relatively strengthened.
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MP. The discussion of different methods and mechanisms for epoxidation is good and generally the introduction is clearly presented. It is impressive that you managed to get the full suite of calculations done. The 13C NMR comparison between computational and experimental data is good and shows the applicability of the method to this problem. These data are often better represented graphically (bar chart of other) to get a feel for the size and spread of error in the calculation. It would also have been interesting to see a comparison between the calculated data for one isomer and both sets of experimental data in order to judge whether the calculation can predict which isomer is which. It is interesting that the 1H NMR calculations came out well in this case because the results of such calculations are not always very accurate. Tools such as the Janocchio calculator based on empirical data are often used to estimate coupling constants which are important for differentiating different isomers (e.g. E an Z double bonds) Compounds 14 and 15 are not enantiomers, they are diastereomers; two compounds with multiple stereocentres are enantiomers if all of the stereocentres have the opposite R/S configuration. For these two compounds only 2 out of 3 stereocentres have different configuration. You didn’t mention how the isomers would be differentiated practically; the best way to do this is X-ray crystallography which shows the exact configuration at all centres. Alternatively this can often be done by NMR experiments (such as diagnostic coupling constants) or 2D experiments (NOESY, HMBC, HSQC) which show the relationship between atoms that are close in space or in number of separating bonds.